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Algebra Notes
Sept. 21: Ideals
Geoffrey Scott
In group theory, the kernels of group homomorphisms weren’t just any subgroups, they were a
special sort of subgroup called a normal subgroup. Likewise, the kernels of ring homomorphisms
will be a special kind of subring called an ideal. To prepare for the definition of an ideal, let’s
quickly review normal subgroups.
Review of Normal Subgroups
Suppose we have a group (G, ?) which is not necessarily abelian.1 Given g ∈ G and a
subgroup H ⊆ G, we can define the left coset g ? H := {g ? h | h ∈ H} and the right coset
H ? g := {h ? g | h ∈ H}. As g varies, these left cosets will partition G, and the right cosets
will also partition G. Now suppose we study the left cosets by asking ourselves the following
question:
Take an element a from a left coset A and multiply it by an element b from
another left coset B. Which left coset will a ? b be inside? Does it depend
on which specific elements from A and B that we picked? If not, then this
recipe defines an operation on the set of left cosets, which would be great!
The sad truth, which we learned in group theory, is that in general the left coset containing
a ? b will depend on exactly which a and b we pick from A and B. For example, take a0 = a ? ha
and b0 = b ? hb , so a0 ∈ A and b0 ∈ B. Then
a0 ? b0 = a ? ha ? b ? hb
and as hard as we might try, we can’t simplify the right hand side to prove that it’s in the same
left coset as ab. However, if we knew that there is some h0a ∈ H such that ha ? b = b ? h0a , then
a0 ? b0 = a ? ha ? b ? hb = a ? b ? (h0a ? hb )
would indeed belong to the same left coset as a ? b. This kind of reasoning leads to following
fact in group theory:
Fact: For a subgroup H of G, the following three statements are equivalent
1. For every h ∈ H and g ∈ G, there is some h0 ∈ H such that hg = gh0 .
2. The partition of G into left cosets of H equals the partition of G into right cosets of
H.
3. The operation [a] ? [b] = [a ? b] on left (and right) cosets is well-defined.
A subgroup H is called normal if the above (equivalent) conditions are true; in this case,
the left cosets are the same as the right cosets (so we just use the word “cosets” to refer to
both), and we call the group of cosets (using the group operation from the third item in the
fact above) the quotient group of G by H, written G/H.
Another way your professor might have motivated the study of normal subgroups is by
proving that for any group homomorphism ϕ : G → G0 , the kernel ker(ϕ) isn’t just any subgroup
1
I am using the ? symbol for the group operation in this lecture to distinguish it from the symbols for addition
or multiplication in a ring.
– it has the nice property that for any g ∈ G and h ∈ ker(ϕ), there is another h0 ∈ ker(ϕ) such
that h ? g = g ? h0 . But this is exactly the property of being a normal subgroup that we saw in
our discussion of “what makes the group operation on cosets well-defined.”
These two ways of motivating the property of normality (as the condition that ensures that
the cosets form a group, or as a curious property of kernels of homomorphisms) may seem
different, but are actually related by the first isomorphism theorem of groups which states that
for any homomorphism ϕ,
G ∼
= Im(ϕ)
ker(ϕ)
This isomorphism is one of the most conceptually important facts in all of group theory; it
should be part of your subconscious intuition about homomorphisms and normal subgroups by
now.
Let’s repeat this same thought experiment in the context of rings instead of groups.
Ideals: the “Normal Subgroups of Ring Theory”
Suppose S is a subring of a ring R. R is an abelian group with respect to the addition
operation, so because every subgroup of an abelian group is normal, S is a normal subgroup of
R. Therefore the operation
[a] + [b] = [a + b]
is well-defined, and R/S is a group. But what about multiplication? Can we multiply cosets
to make R/S into a ring? To answer this, we first consider the following question
Take an element a from a coset A and an element b from another coset B.
We already know that if we add a and b, the coset containing a + b doesn’t
depend on which specific elements from A and B we picked. But now what
if we multiply the elements a and b? Will the coset of ab depend on which
elements that we chose? If not, then the operation
[a][b] = [ab]
would be well-defined!
The sad truth is that in general the coset containing ab will depend on exactly which a and b
we pick from A and B. For example, take a0 = a + sa and b0 = b + sb , so a0 ∈ A and b0 ∈ B
(remember, the group operation that we’re using to define the cosets is addition – this is why
we’ve written a0 = a + sa not a0 = asa ). Then
a0 b0 = ab + asb + sa b + sa sb
If we knew that asb ∈ S and sa b ∈ S, then a0 b0 would be in the same coset as ab. One condition
on the subring S that causes this property to be true leads to the definition of an ideal.
Definition: Let S be a subring of R. If, for any elements s ∈ S and r ∈ R, we
have rs ∈ S and sr ∈ S, then S is an ideal of R.
Notice that the definition of a subring already guarantees that if s, r ∈ S, then rs and sr
are in S. The condition of being an ideal is stronger: it says that when you multiply an element
of S by any element (even one outside of S), the result is inside S. You can think that an ideal
“absorbs” the ring into the ideal. Check that the following are ideals:
Example 0: In any ring R, the zero ideal (consisting of just the element 0) and the whole ring
R are both ideals of R.
Example 1: For any n ∈ Z, the subset nZ ⊆ Z is an ideal of Z
Example 2: In R[x], the subring of polynomials that have zero constant term is an ideal.
Example 3: In C(R), the set of continuous functions that vanish at x = 9 is an ideal.
Relationship between ideals and quotient rings
Proposition: If I ⊆ R is an ideal, then R/I is a ring, with addition and multiplication of
cosets defined by
[a] + [b] := [a + b]
and
[a][b] := [ab]
Proof: The discussion above proves that the addition and multiplication operations are welldefined. Also, because I ⊆ R is a normal subgroup, we know from group theory that R/I
is a group with respect to the addition operation defined above. It is abelian because
[a] + [b] = [a + b] = [b + a] = [b] + [a]
The fact that multiplication is associative is shown by
[a]([b][c]) = [a][bc] = [a(bc)] = [(ab)c] = [ab][c] = ([a][b])[c]
and the fact that distribution holds is given by
[a]([b] + [c]) = [a][b + c] = [a(b + c)] = [ab + ac] = [a][b] + [a][c]
(the fact that ([b] + [c])[a] = [b][a] + [c][a] is proved in a similar way).
Proposition: Let ϕ : R → S be a ring homomorphism. Then ker(ϕ) is an ideal in R, and
R ∼
= Im(ϕ)
ker(ϕ)
Proof: We already proved that ker(ϕ) is a subring; now, we must verify that it is an ideal. Let
r ∈ R and k ∈ ker(ϕ). The fact that kr, rk ∈ ker(ϕ) is shown by
ϕ(rk) = ϕ(r)ϕ(k) = ϕ(r)(0) = 0
ϕ(kr) = ϕ(k)ϕ(r) = (0)ϕ(r) = 0.
To prove the isomorphism, define the map
R
→ Im(ϕ)
ker(ϕ)
[r] 7→ ϕ(r)
This is well-defined, since ϕ(r + k) = ϕ(r) + ϕ(k) = ϕ(r) for any k ∈ ker(ϕ).
The kernel of this map is the set of cosets [r] for which ϕ(r) = 0. This is just the zero
coset, so the kernel is zero and therefore the map is injective.
The fact that the map is surjective is seen by taking an arbitrary element s of Im(ϕ), so
s = ϕ(r) for some r ∈ R, and observing that [r] is sent to ϕ(r) = s under the map above.
Therefore, the above map is injective and surjective, hence an isomorphism.
Question: Determine whether the following subsets of Z × Z are subrings, ideals, or neither.
If it is an ideal I, describe (Z × Z)/I.
• {(a, a) | a ∈ Z}.
• {(2a, 2b) | a, b ∈ Z}.
• {(2a, 0) | a ∈ Z}.
• {(a, −a) | a ∈ Z}.
Question: Determine whether the following subsets of Z[x] are subrings, ideals, or neither. If
it is an ideal I, describe Z[x]/I.
• All polynomials whose constant term is a multiple of 5.
• All polynomials whose coefficient of x5 is 5.
• All polynomials whose constant term is 0, and whose coefficient of x is 0.
• All polynomials p such that p(7) = 0.
• All polynomials p such that p(7) = 1.
• All polynomials where only even powers of x appear (this is sometimes written Z[x2 ]).
• All polynomials whose coefficients sum to zero.
• All polynomials whose coefficients are divisible by 3.
Properties of ideals: a preview
Next class, we will learn about all sorts of properties of ideals. Here’s one today.
Proposition: Let I be an ideal in a ring R. If I contains a unit, then I = R.
Proof: Let u ∈ I be a unit. This means that u has a multiplicative inverse u−1 ∈ R. Then for
any a ∈ R, we see that a = au−1 u, so a ∈ I.
Corollary: The only ideals in a field F are I = {0} and I = R.